小白笔记-----------------------leetcode(8. String to Integer (atoi) )

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Requirements for atoi:

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

开始想用，parseInt 和trim()解决问题，后来发现不行，

首先第一要判断是否为空，为空返回

第二步，可以利用trim()先解决空格问题，然后判断正负值

第三步，遍历每个值，查看是否不是0-9的值，进行求和

第四步，判断是否超过最大，最小值

这里有个细节那就是结果的数据类型，要用double类型

代码：

public class Solution {      public int myAtoi(String str) {        int i = 0;        if(str.length() <  1){            return 0;        }        str = str.trim();        int sign = 1;        if(str.charAt(0) == '+'){            i++;        }else if(str.charAt(0) == '-'){            sign = -1;            i++;        }        double result = 0;        while(i < str.length() && str.charAt(i) >= '0' && str.charAt(i) <= '9'){            result = result * 10 + (str.charAt(i) - '0');            i++;        }        if(sign == -1){            result = -result;        }        if(result > Integer.MAX_VALUE){            return Integer.MAX_VALUE;        }        if(result < Integer.MIN_VALUE){            return Integer.MIN_VALUE;        }                return (int)result;    }}